I'm guessing you're a physicist by your unusual notation! The real-analytic Eisenstein series is usually denoted $E(z,s)$ (where $z = \tau$ in your notation), not $E_s(\tau,\bar{\tau})$ (why does $\bar{\tau}$ appear?). I'll also let $f(z)$ denote a cusp form rather than $\nu_i(\tau,\bar{\tau})$ (I've never seen this notation used).

Anyway, the question asks (among other things) the following. Given two cusp forms $f_1,f_2$ and an Eisenstein series $E(z,s)$, determine a closed formula for
$$\int_{\Gamma \backslash \mathbb{H}} f_1(z) f_2(z) E(z,s) \, d\mu(z).$$
Here $\Gamma = \mathrm{SL}_2(\mathbb{Z})$ and $d\mu(z) = y^{-2} \, dx \, dy$.

This question has a nice answer, as does the same question where one allows any triple product involving various cusp forms and Eisenstein series. When at least one form is an Eisenstein series, this is the Rankin-Selberg method; when all three are cusp forms, this is known as the Watson-Ichino triple product formula. In Section 2.2 of this paper of mine, I write out the exact formulae for these in several cases, and provide some references.

Let me give a more detailed explanation of this in the case written above. It is known, going back to work of Hecke, that any Maass cusp form $f$ can be written as a finite linear combination of Hecke-Maass cusp forms; these are eigenfunctions of the Hecke operators and have Fourier expansions of the form
$$f(z) = \rho_f \sum_{n = 1}^{\infty} \frac{\lambda_f(n)}{\sqrt{n}} W_{0,it_f}(4\pi ny) (e^{2\pi i nx} + \epsilon_f e^{-2\pi i nx}).$$
Here $\rho_f$ is a normalising constant, $\lambda_f(n)$ is the $n$-th Hecke eigenvalue (which is multiplicative as an arithmetic function and conjecturally satisfies $\lambda_f(p) \in [-2,2]$ for any prime $p$), $t_f$ is the spectral parameter of $f$ (so that $f$ has Laplacian eigenvalue $\lambda_f = 1/4 + t_f^2$), $W_{\alpha,\beta}$ denotes the Whittaker function (in this case, it is basically just a $K$-Bessel function), and $\epsilon_f \in \{1,-1\}$ is the parity of $f$ (so that $f(-\overline{z}) = \epsilon_f f(z)$).

So we may assume that $f_1$ and $f_2$ are of this form. One then *unfolds* the integral by inserting the identity
$$E(z,s) = \sum_{\gamma \in \Gamma_{\infty} \backslash \Gamma} \Im(\gamma z)^s,$$
using the fact that $f_1(\gamma z) f_2(\gamma z) = f_1(z) f_2(z)$, and that $\Gamma_{\infty} \backslash \mathbb{H} \cong \{z : x \in [0,1], \ y > 0\}$. One then inserts the Fourier expansions of $f_1$ and $f_2$ and evaluates the $x$ integral, then the $y$ integral; this latter integral is essentially the Mellin transform of the product of two $K$-Bessel functions, which has an explicit expression in terms of products of gamma functions.

All in all, one finds that the triple product is equal to
$$\frac{\rho_{f_1} \rho_{f_2} \pi^{-s}}{\Gamma(s)} \prod_{\pm_1, \pm_2} \Gamma\left(\frac{s \pm_1 it_{f_1} \pm_2 it_{f_2}}{2}\right) \sum_{n = 1}^{\infty} \frac{\lambda_{f_1}(n) \lambda_{f_2}(n)}{n^s}.$$
(At least if $\epsilon_{f_1} = \epsilon_{f_2}$; otherwise, the gamma functions are changed slightly.) Here we are initially assuming that $\Re(s) > 1$; however, by analytic continuation, this extends meromorphically to all of $\mathbb{C}$ (with a simple pole at $s = 1$ if $f_1 = f_2$ and no poles otherwise). Let $\zeta(s) = \sum_{n = 1}^{\infty} n^{-s}$ denote the Riemann zeta function. Then we *define* the Rankin-Selberg $L$-function $L(s,f_1 \times f_2)$ to be
$$L(s,f_1 \times f_2) = \zeta(2s) \sum_{n = 1}^{\infty} \frac{\lambda_{f_1}(n) \lambda_{f_2}(n)}{n^s}.$$
Again, this is initially defined for $\Re(s) > 1$ in this way, but extends meromorphically to $\mathbb{C}$. Moreover, this has a nice functional equation, which can be proven using the functional equation for $E(z,s)$.

Next, let me discuss the same problem for
$$\int_{\Gamma \backslash \mathbb{H}} f(z) E\left(z,\frac{1}{2} + it\right) E(z,s) \, d\mu(z).$$
The key point is that $E(z,1/2 + it)$ has a nice Fourier expansion; $t$ replaces $t_f$, while $\lambda_f(n)$ is replaced by $\lambda(n,t) = \sum_{ab = n} a^{it} b^{-it}$. The exact same method works, except that the ensuing Rankin-Selberg $L$-function factorises as the product of two $L$-functions:
$$\zeta(2s) \sum_{n = 1}^{\infty} \frac{\lambda_f(n) \lambda(n,t)}{n^s} = \sum_{m = 1}^{\infty} \frac{\lambda_f(m)}{m^{s + it}} \sum_{n = 1}^{\infty} \frac{\lambda_f(n)}{n^{s - it}} = L(s + it,f) L(s - it,f).$$

Now we consider
$$\int_{\Gamma \backslash \mathbb{H}}^{\mathrm{reg}} E\left(z,\frac{1}{2} + it_1\right) E\left(z,\frac{1}{2} + it_2\right) E(z,s) \, d\mu(z).$$
I have written $\int^{\mathrm{reg}}$ to denote the fact that we must *regularise* (or *renormalise*) this integral in order for it to converge. This goes back to work of Zagier. Here not only does the Rankin-Selberg $L$-function factorise into the product of two $L$-functions, but each of these $L$-functions factorises further into the product of two copies of the Riemann zeta function.

Finally, let me consider
$$\int_{\Gamma \backslash \mathbb{H}} f_1(z) f_2(z) f_3(z) \, d\mu(z).$$
Here the Rankin-Selberg method does not apply, since we cannot unfold as there is no Eisenstein series present. Nonetheless, this triple product can be shown (via methods coming from representation theory) to be associated to $L$-functions. This was done in Watson's thesis and was further expanded upon by Ichino. What they show is an identity relating the *square of the absolute value* of this integral to a certain $L$-function, namely $L(1/2, f_1 \times f_2 \times f_3)$.